Vtyjkyuk

Problem: Let $bin E$(normed vector space), show that $d(b,B(a,r))=d(b,overlineB(a,r))$.

Notation: $d(a,C)=infd(a,x):xin C$

My attempt: Well, since $B(a,r)subset overline(B)(a,r)$, then $d(b,overlineB(a,r)) leq d(b,B(a,r))$. So, I suppose $d(b,overlineB(a,r)) < d(b,B(a,r))$ to find a contradiction.

The real problem is that I have no idea how to continue nor even if im in the right path.

PD: Please do not use convergence :)

Thanks for the help!

Let $x in overset - B (a,r)$. Then there exists $x_n subset B (a,r)$ such that $x_n to x$. We have $d(b,B (a,r)) leq d(b,x_n)$ for all $n$. If you let $n to infty$ you get $d(b,B (a,r)) leq d(b,x)$. Taking inf over all $x$ we get $d(b,B (a,r)) leq d(overset - B (a,r))$. For a different proof not involving sequences let $x in overset - B (a,r)$ and $epsilon >0$. There exists $y in B(b,r)$ such that $d(x,y) <epsilon$. We have $d(b,B(a,r))leq d(b,y)<d(b,x)+epsilon$. Take infimum over $x$ to get $d(b,B(a,r))leq d(b,overset - B (a,r))+epsilon$. Since $epsilon$ is arbitrary this gives $d(b,B(a,r))leq d(b,overset - B (a,r))$

A normed vector space $E$ is also a metric space with metric $d(x,y)=|x-y|.$ In a metric space $(E,d)$ let $emptyset ne Csubset E$ and $bin E.$ We define $d(b,C)=inf d(b,c):cin C.$ We have $d(b,c'):c'in bar Csupset d(b,c):cin C$ so $d(b, bar C)leq d(b,C).$

We show that $d(b,bar C)<d(b,C)$ is untenable, and conclude that $d(b,bar C)=d(b,C).$

For brevity of notation let $U=d(b,C)$ and $V=d(b,bar C).$ Suppose $ U-V>0.$ Take $c'in bar C$ such that $d(b,c')leq V+frac U-V2.$ Now $forall cin C;(d(c,b)geq U).$ So for every $cin C$ we have $$d(c,c')geq d(c,b)-d(b,c')geq$$ $$geq U-(V+frac U-V2)=frac U-V2. $$ So the open ball of positive radius $frac U-V2,$ centered at $c',$ is disjoint from $C.$ This contradicts $c'in bar C. $